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Consider an integrator that has the input-output relation

Determine the input-output relation for the inverse system.

The first-order difference equation y[n] - ay[n - 1] = x[n], 0 < a < 1, describes a particular discretetime system initially at rest. (a) Verify that the impulse response h[n] for this system is h[n] = a'u[n]. (b) (b) Is the system (i) memoryless? (ii) causal? (iii) stable?

Clearly state your reasoning. (c) Is this system stable if |a I > 1?

The first-order differential equation dy(t) + 2y(t) = x(t) dt describes a particular continuous-time system initially at rest.

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(a) Verify that the impulse response of this system is h(t) = e - 2 u(t). (b) (b) Is this system (i) memoryless? (ii) causal? (iii) stable?

Clearly state your reasoning Consider the linear, time-invariant system in Figure P5.4, which is composed of a cascade of two LTI systems. u(t) is a unit step signal and s(t) is the step response of system L.

Using the fact that the overall response of LTI systems in cascade is independent of the order in which they are cascaded, show that the impulse response of system L is the derivative of its step response, i.e.,

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Consider the cascade of two systems shown in Figure P5.5. System B is the inverse of system A.

(a) Suppose the input is b(t). What is the output w(t)? (b) Suppose the input is some more general signal x(t). What is the output w(t) in terms of x(t) (a) Consider again the cascade of two systems presented in Problem P5.5. Suppose an input x1(t) produces yi(t) as system A output and an input x 2(t) produces y2(t) as system A output. What is w(t) if the input is such that y(t), the output of system A, is ay 1(t) + by 2(t) with a, b constants? (b) Suppose an input x 1(t) produces yi(t) as system A output. What is w(t) if x(t) is such that y(t) = y 1(t - r)? (c) (c) Is system B an LTI system? Justify your answer.

Consider the three discrete-time signals shown in Figure P5.7.

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(a) Verify the distributive law of convolution: (x + w) * y = (x * y) + (w * y) (b) You may have noticed a similarity between the convolution operation and multiplication, but they are not equivalent. Verify that (x * y) - w # x * (y - w)

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Determine if each of the following statements concerning LTI systems is true or false. Justify your answers. (a) If h(t) is the impulse response of an LTI system and h(t) is periodic and nonzero, the system is unstable. (b) The inverse of a causal LTI system is always causal. (c) If Ih[n] 5 K for each n, where K is a given number, then the LTI system with h[n] as its impulse response is stable. (d) If a discrete-time LTI system has an impulse response h[n] of finite duration, the system is stable. (e) If an LTI system is causal, it is stable. (f) The cascade of a noncausal LTI system with a causal one is necessarily noncausal. (g) A continuous-time LTI system is stable if and only if its step response s(t) is absolutely integrable, i.e.,

(h) A discrete-time LTI system is causal if and only if its step response s[n] is zero for n < 0.

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In Section 3.7 of the text we characterized the unit doublet through the equation

for any signal x(t). From this equation we derived the fact that

(a) Show that eq. (P5.10-2) is an equivalent characterization of ui(t) by showing that eq. (P5.10-2) implies eq. (P5.10-1). [Hint:Fix t and define the signal g(r) x(t - r).]

Thus we have seen that characterizing the unit impulse or unit doublet by how it behaves under convolution is equivalent to characterizing how it behaves under integration when multiplied by an arbitrary signal g(t). In fact, as indicated in Section 3.7 of the text, the equivalence of these operational definitions holds for all signals and in particular for all singularity functions. (b) Let f(t) be a given signal. Show that

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by showing that both have the same operational definitions. (c) Determine the value of

(d) Find an expression for f(t)u2 (t) analogous to that considered in part (b) for f(t)ui(t) Consider the cascade of two systems H and G as shown in Figure P5.11.

(a) If H and G are both LTI causal systems, prove that the overall system is causal. (b) (b) If H and G are both stable systems, show that the overall system is stable. (c) Find the combined impulse response of the LTI system in Figure P5.12. Recall that x(t)* h(t) * h1(t) = x(t).

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Find the necessary and sufficient condition on the impulse response h[n] such that for any input x[n], max {Ix[n]I} > max {Iy[n]|}, where y[n] = x[n] * h[n].

The inverse system for a continuous-time accumulation (or integration) is a differentiator. This can be verified because

Therefore, the input-output relation for the inverse system in Figure S5.1 is

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(a) We want to show that (b) h[n] - ah[n - 11 = b[n] (c) Substituting h[n] = anu[n], we have (d) anu[n] - aa"-u[n - 1] = a"(u[n] - u[n - 1]) But u[n] - u[n - 1] = b[n] and an6[n] = ab[n] = b[n] (i) The system is not memoryless since h[n] # kb[n]. (ii) The system is causal since h[n] = 0 for n < 0. (iii) The system is stable for Ia I < 1 since

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is bounded. (c) The system is not stable for |al > 1 since E"o 1a|" is not finite. (a) Consider x(t) = 6(t) -+ y(t) = h(t). We want to verify that h(t) = e -2u(t), so

but ed' b(t) = 6(t) because both functions have the same effect on a test function within an integral. Therefore, the impulse response is verified to be correct. (b) (i) The system is not memoryless since h(t) # kb(t). (ii) The system is causal since h(t) = 0 for t < 0. (iii) The system is stable since h(t) is absolutely integrable.

By using the commutative property of convolution we can exchange the two systems to yield the system in Figure S5.4.

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Now we note that the input to system L is

so y(t) is the impulse response of system L. From the original diagram,

Therefore,

(a) By definition, an inverse system cascaded with the original system is the identity system, which has an impulse response h(t) = 6(t). Therefore, if the cascaded system has an input of b(t), the output w(t) = h(t) = 6(t). (b) Because the system is an identity system, an input of x(t) produces an output w(t) = x(t).

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(a) If y(t) = ayi(t) + by 2(t), we know that since system A is linear, x(t) = ax,(t) + bx 2 (t). Since the cascaded system is an identity system, the output w(t) = ax1(t) + bx 2(t).

(b) If y(t) = y 1(t - r), then since system A is time-invariant, x(t) = x,(t - -) and also w(t) = xi(t - r).

(c) From the solutions to parts (a) and (b), we see that system B is linear and time invariant (a) The following signals are obtained by addition and graphical convolution:

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Therefore, the distributive property (x + w) * y = x * y + w * y is verified. (b) Figure S5.7-3 shows the required convolutions and multiplications.

Note, therefore, that (x[n] * y[n]) - w[n] # x~n] * (y[n] - w[n]).

Consider

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where the primes denote d/dt.

True

(b) False. If h(t) = 3(t - to) for to > 0, then the inverse system impulse response is b(t + to), which is noncausal. (c) False. Suppose h[n] = u[n]. Then

(d) True, assuming h[n] is finite-amplitude.

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(e) False. h(t) = u(t) implies causality, but J u(t) dt = oo implies that the system is not stable. (f) False.

(g) False. Suppose h(t) = e-'u(t). Then

The step response is

(h) True. We know that u[n] = E=O b[n - k] and, from superposition, s[n] = Ef=0 h[n - k]. If s[n] # 0 for some n < 0, there exists some value of h[k] # 0 for some k < 0. If s[n] = 0 for all n < 0, h[k] = 0 for all k < 0.

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Noting that 2g'(7r)f'(r) ,= = - 2f'(O) fg(r)ui(r)dr, we have an equivalent operational definition: f(r)u 2(r) = f(O)u 2(r) - 2f'(0)ui(r) + f"(O)b(r) (a) h(t) * g(t) = J'. h(t - r)g(r) dr = f' h(t - )g(r) dr since h(t) = 0 for t < 0 and g(t) = 0 for t < 0. But if t < 0, this integral is obviously zero. Therefore, the cascaded system is causal. (b) By the definition of stability we know that for any bounded input to H, the output of H is also bounded. This output is also the input to system G. Since the input to G is bounded and G is stable, the output of G is bounded. Therefore, a bounded input to the cascaded system produces a bounded output. Hence, this system is stable.

We have a total system response of

We are given that y[n] = x[n] * h[n].

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We can see from the inequality

that E Ih[k]| 5 1 max {Iy[n]|} s max {Ix[n]|}. This means that E Ih[k]I - 1 is a sufficient condition.

It is necessary because some x[n] always exists that yields y[n] = E=_ Ih[k]1. (x[n] consists of a sequence of +1's and -l's.) Therefore, since max {x[n]} = 1, it is necessary that E= h[k] 1 5 1 to ensure that y[n] S max {Ix[n]I} = 1.